# What is the slope of the tangent line of  x^2/(x-y^3) = C , where C is an arbitrary constant, at (-2,1)?

Jan 2, 2016

$- \frac{3}{2}$

#### Explanation:

Find the implicit derivative of the function.

Use the product rule.

$\frac{\left(x - {y}^{3}\right) \frac{d}{\mathrm{dx}} \left({x}^{2}\right) - {x}^{2} \frac{d}{\mathrm{dx}} \left(x - {y}^{3}\right)}{x - {y}^{3}} ^ 2 = 0$

Remember that differentiating a $y$ term will spit out a $\frac{\mathrm{dy}}{\mathrm{dx}}$ term thanks to the chain rule.

$\frac{2 x \left(x - {y}^{3}\right) - {x}^{2} \left(1 - 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}\right)}{x - {y}^{3}} ^ 2 = 0$

Plug in $\left(- 2 , 1\right)$ for $\left(x , y\right)$ and solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$.

$\frac{2 \left(- 2\right) \left(- 2 - {1}^{3}\right) - {\left(- 2\right)}^{2} \left(1 - 3 {\left(1\right)}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}\right)}{- 2 - {1}^{3}} ^ 2 = 0$

Notice that the denominator can be cancelled out. Continue simplification of numerator.

$\left(- 4\right) \left(- 3\right) - \left(4\right) \left(- 2\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$12 + 8 \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3}{2}$

The slope of the tangent line is $- \frac{3}{2}$.